Solving the Polynomial Equation x4 - 4x3 - 1 0: A Comprehensive Guide

In this guide, we will explore how to solve the polynomial equation x4 - 4x3 - 1 0 using the rational root theorem and the Newton method. We'll also consider the numerical solution and the application of Vieta's relationships to find the roots.

Rational Root Theorem Application

First, let's consider the polynomial equation x4 - 4x3 - 1 0. By the rational root theorem, the only possible rational roots are x 1 and x -1. However, substituting these values into the equation, we find:

For x 1, we get:

14 - 4(13) - 1 1 - 4 - 1 -4 ≠ 0

For x -1, we get:

(-1)4 - 4(-13) - 1 1 4 - 1 4 ≠ 0

Thus, there are no rational roots, and no factors of the form (x - a) or (bx - c) for integers a and b.

Numerical Solution with Near Miss Function

We can solve this polynomial equation numerically by considering a near miss function. Let us define:

$f(x) x4 - 4x3 - 1$

Consider the near miss function:

$g(x) x4 - 4x3$

The roots of this near miss function are 0, 0, 0, and 4. This suggests that the roots of the original polynomial are near 0 and 4. Evaluating f(0) and f(4), we find:

$f(0) -1$ and $f(4) -1$, which are close to zero.

Newton Method Application

To find the roots, we apply the Newton method. The Newton method for finding the root of a function $f(x)$ is given by:

$x_{n 1} x_n - frac{f(x_n)}{f'(x_n)}$

First, let's compute the derivative of f(x):

$f'(x) 4x^3 - 12x^2$

Starting at x 4, the iterations are:

$x_1 4 - frac{f(4)}{f'(4)} 4 - frac{-1}{4(4^3) - 12(4^2)} 4 - frac{-1}{256 - 192} 4 - frac{-1}{64} 4.015625$

For x 0, the iterations are:

$x_1 0 - frac{f(0)}{f'(0)} 0 - frac{-1}{4(0^3) - 12(0^2)} 0 - frac{-1}{0} -0.60123$

The Newton method converges to these values.

Vieta's Relationships and Final Roots

Using Vieta's relationships, we can find the remaining roots. Let r, s, t, u be the roots of the polynomial such that:

$r s t u 4.0156$

$rs rt ru st su tu 0$

$rst rsu rtu stu -1$

$rstu -1$

We know:

$r 4.0156$ and $s -0.60123$

Using these, we can calculate tu:

$tu 4 - rs 4 - 3.4147 0.5853$

$tu 0.41422$

Thus, the remaining roots are the solutions to:

$x^2 - 0.5853x 0.41422 0$

The roots of this quadratic equation are:

$t 0.29265 - 0.573215i$

$u 0.29265 0.573215i$

Hence, the polynomial x4 - 4x3 - 1 0 has the roots 4.0156, -0.60123, 0.29265 - 0.573215i, and 0.29265 0.573215i.